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Question

A long straight wire of radius a carries a steady current I. The current is uniformly distributed across its cross-section. The ratio of the magnetic field at distance a2 and 2a from axis of wire is

A
12
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B
14
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C
4
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D
1
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Solution

The correct option is D 1
Here, current is uniformly distributed across the cross-section of the wire, thus current per unit cross-sectional area is

iA=Iπa2

Case 1: B at a distance r=a2

The current enclosed by loop 1 is,


i1=iA×π(a2)2

or, i1=Iπa2×πa24

or, i1=I4

Using Ampere circuital law on loop 1 we have,

B1.dl=μ0Ien

B1dl=μ0I4 (θ=0,Ienc=I4)

B1(2πa2)=μ0(I4)

or, B1=μ0I4πa

Case 2: B at a distance r=a2

Similarly, for loop 2, r=2a

B2.dl=μ0Ien

B2(2π×2a)=μ0I (Ien=I)

B2=μ0I4πa

B1B2=1
Why this Question ?
Key point- Magnetic field calculation with the help of Ampere circuital law is easily possible only under highly symmetric conditions.
Ampere circuital law is given by,

B.dl=μ0Ien

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